But you're certainly on the right path to solving this. You know\[\int\limits_{0}^{5}[f(x)+g(x)] dx=12\]So you just need to find \[\int\limits_{-5}^{0}[f(x)+g(x)]dx =\int\limits_{-5}^0f(x)\;dx+\int\limits_{-5}^0g(x)\;dx\]First, lets start with the f(x) part.

Since \(f(x)=f(-x)\), we have that \[\begin{aligned}
\int\limits_{-5}^0f(x)\;dx&=\int\limits_{-5}^0f(x)\;dx \\
&=\int\limits_{-5}^0f(-x)\;dx\\
&=\int\limits_{5}^0f(-u)\;(-du)\qquad\text{this is a u-sub for }u=-x.\\
&=\int\limits_0^5f(-u)\;du\\
&=\int\limits_0^5f(u)\;du\\
&=8
\end{aligned}\]

This is basically the same thing you do for \(g(x)\). Instead we have \(-g(x)=g(-x)\).If we repeat the above argument, we get to the point \[\int\limits_0^5g(-u)\;du\]Substitute for \(-g(u)\), and we get \[-\int\limits_0^5g(u)\;du=-4\]